Help someone please!. What is integral of (t raised by power 1/3) all multiplied by sin(2(t squared)-3t) or using the ^ symbol to indicate powers and * to indicate multiplication, what is the integral of t^(1/3) * sin(2(t^2)-3t). This is a past exam question I'm having trouble with. I've tried 'by parts' but can't get anywhere. Sorry to anyone who was investigating my original wording of this. I mixed variables here ... everything should have been x or t ... I've now changed to t as per the original question.

Question

Help someone please!. What is integral of  (t raised by power 1/3) all multiplied by sin(2(t squared)-3t) or using the ^ symbol to indicate powers and * to indicate multiplication, what is the integral of  t^(1/3) * sin(2(t^2)-3t). This is a past exam question I'm having trouble with. I've tried 'by parts' but can't get anywhere. Sorry to anyone who was investigating my original wording of this. I mixed variables here ... everything should have been x or t ... I've now changed to t as per the original question.

anonymous · Answer

what is t u can only integrate with respect to one variable either t or x

anonymous · Answer

Sorry, my mistake which I had just noticed. Both variables should have been t.

agent0smith · Answer

$$\Large \int\limits_{}^{} t^{1/3} \sin(2t^2 - 3t) dt =$$

anonymous · Answer

Yes, that's the problem. I'm a newbie and didn't realise there is an equation editor to present the problem more clearly. Thanks.

agent0smith · Answer

Well... wolfram alpha doesn't even post a result with normal computation time, so i have no idea how to go about integrating it: http://www.wolframalpha.com/input/?i=integral+t%5E%7B1%2F3%7D+%5Csin%282t%5E2+-+3t%29+dt+

anonymous · Answer

Okay, thanks anyway. I've tried 'by parts' and 'by substitution'. I'm beginning to suspect it was a mistake on the examiner's part - i.e. that he did not mean to ask this question.

anonymous · Answer

i think u have copied the question wrong

anonymous · Answer

Hi Chandanjha, I entered the question incorrectly initially and not only that I hate to admit I even had an error in my first correction of it. Sorry, I'm feeling under pressure here with a repeat exam imminent. The correct version, straight from the original exam paper, and I've double checked it is:
 $$\int\limits(t^{\frac{ 2 }{ 3}})\sin(t ^{2}-3t)dt$$
Apologies again to agent0smith. I had a constant multiplier of 2 on the t-squared term inside the sine in my earlier incorrect version. However, I don't think this corrected version makes it any easier.

anonymous · Answer

can u draw the integral

agent0smith · Answer

http://www.wolframalpha.com/input/?i=%5Cint+%28t%5E%7B%5Cfrac%7B+2+%7D%7B+3%7D%7D%29%5Csin%28t+%5E%7B2%7D-3t%29dt

anonymous · Answer

I'm beginning to feel really stupid here. I left out the limits (upper limit = $$x ^{2}$$ and lower limit = 2) and the first power is 1/3 on the t. Sorry, did not get any sleep last night. Here is the drawing.  |dw:1375099176175:dw|